Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 3 Algebra Intext Questions

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 3 Algebra Intext Questions

Exercise 3.1

Try These (Text book Page No. 44)

Question 1.
Observe and complete the following table. First one is done for you.

Solution:

Try These (Text book Page No. 46)

Question 1.
Simplify and write the following in exponential form.
1. 2³ × 2⁵
2. p² × P⁴
3. x⁶ × x⁴
4. 3¹ × 3⁵ × 3⁴
5. (-1)² × (-1)³ × (-1)⁵
Solution:
1. 2³ × 2⁵ = 2³⁺⁵ = 2⁸ [since a^m × aⁿ = a^m+n]

2. p² × p⁴ = p²⁺⁴ = p⁶ [since a^m × aⁿ = a^m+n]

3. x⁶ × x⁴ = x⁶ + 4 = x¹⁰ [since a^m × aⁿ = a^m+n]

4. 3¹ × 3⁵ × 3⁴ = 3¹⁺⁵ × 3⁴ [since a^m × aⁿ = a^m+n]
= 3⁶ × 3⁴ [since a^m × aⁿ = a^m+n]
= 3¹⁰

5. (-1)² × (-1)³ × (-1)⁵
= (-1)²⁺³ × (-1)⁵ [Since a^m × aⁿ = a^m+n]
= (-1)⁵ × (-1)⁵
= (-1)⁵⁺⁵ [Since a^m × aⁿ = a^m+n]
= (-1)¹⁰

Try These (Text book Page No. 48)

Question 1.
Simply the following.
1. 23⁵ ÷ 23²
2. 11⁶ ÷ 11³
3. (-5)³ ÷ (-5)²
4. 7³ ÷ 7³
5. 15⁴ ÷ 15
Solution:

Try These (Text book Page No. 48)

Question 1.
Simplify and write the following in exponent form.
1. (3²)³
2. [(-5)³]²
3. (20⁶)²
4. (10³)⁵
Solution:
1. (3²)³ = 3^2×3 = 3⁶ [since (a^m)ⁿ = a^m×n]
2. [(-5)]² = (-5)^3×2 = (-5)⁶ [since (a^m)ⁿ = a^m×n]
3. (20⁶)² = 20^6×2 = 20¹² [since (a^m)ⁿ = a^m×n]
4. (10³)⁵ = 10^3×5 = 10¹⁵ [since (a^m)ⁿ = a^m×n]

Question 2.
Express the following exponent numbers using a^m × b^m = (a × b)^m.
(i) 5² × 3²
(ii) x³ × y³
(iii) 7⁴ × 8⁴
Solution:
(i) 5² × 3² = (5 × 3)² = 15² [since a^m × b^m = (a × b)^m]
(ii) x³ × y³ = (x × y)³ = (x y)³
(iii) 7⁴ × 8⁴ = (7 × 8)⁴ = 56⁴

Question 3.
Simplify the following exponent numbers by using (\(\frac { a }{ b } \))^m = \(\frac{a^{m}}{b^{m}}\)
(i) 5³ ÷ 2³
(ii) (-2)⁴ ÷ 3⁴
(iii) 8⁶ ÷ 5⁶
(iv) 6³ ÷ (-7)³
Solution:
(i) 5³ ÷ 2³ = (\(\frac { 5 }{ 2 } \))³ – [Since \(\frac{a^{m}}{b^{m}}\) = (\(\frac { a }{ b } \))^m]
(ii) (-2)⁴ ÷ 3⁴ = (\(\frac { -2 }{ 3 } \))⁴
(iii) 8⁶ ÷ 5⁶ = (\(\frac { 8 }{ 6 } \))⁶
(iv) 6³ ÷ (-7)³ = (\(\frac { 6 }{ -7 } \))³

Exercise 3.2

Try These (Text book Page No. 54)

Question 1.
Find the unit digit of the following exponential numbers:
(i) 106²¹
(ii) 25⁸
(iii) 31¹⁸
(iv) 20¹⁰
Solution:
(i) 106²¹ Unit digit of base 106 is 6 and the power is 21 and is positive.
Thus the unit digit of 106²¹ is 6.

(ii) 25⁸ Unit digit of base 25 is 5 and the power is 8 and is positive.
Thus the unit digit of 25⁸ is 5.

(iii) 31¹⁸ Unit digit of base 31 is 1 and the power 18 and is positive.
Thus the unit digit of 31¹⁸ is 1.

(iv) 20¹⁰ Unit digit of base 20 is 0 and the power 10 and is positive.
Thus the unit digit of 20¹⁰ is 0.

Try These (Text book Page No. 55)

Question 1.
Find the unit digit of the following exponential numbers:
(i) 64¹¹
(ii) 29¹⁸
(iii) 79¹⁹
(iv) 104³²
Solution:
(i) 64¹¹ Unit digit of base 64 is 4 and the power is 11 (odd power).
∴ Unit digit of 64¹¹ is 4.

(ii) 29¹⁸ Unit digit of base 29 is 9 and the power is 18 (even power).
Therefore, unit digit of 29¹⁸ is 1.

(iii) 79¹⁹ Unit digit of base 79 is 9 and the power is 19 (odd power).
Therefore, unit digit of 7919 is 9.

(iv) 104³² Unit digit of base 104 is 4 and the power is 32 (even power).
Therefore, unit digit of 104³² is 6.

Exercise 3.3

Try These (Text book Page No. 35)

Question 1.
Complete the following table:

Solution:

Question 2.
Identify the like terms from the following:
(i) 2x²y, 2xy²,3xy²,14x²y, 7yx
(ii) 3x³y², y³x, y³x², – y³x, 3y³x
(iii) 11pq, -pq, 11pqr, -11pq,pq
Solution:
(i) 2x²y, 2xy², 3xy², 14x²y, 7yx
(a) 2x²y and 14x²y are like terms.
(b) 2xy² and 3xy² are like terms.

(ii) 3x³y², y³x, y³x², – y³x, 3y³x
(a) y³x, – y³x and 3y³x are like terms.

(iii) 11 pq, -pq, 11pqr , -11 pq, pq
(a) 11 pq, -pq, -pq and pq are like terms.